A projectile is shot from the ground at an angle of #pi/12 # and a speed of #8 /15 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

We apply the equation of motion

to calculate the time to reach the greatest height

We apply the equation of motion

To find the distance from the starting point when the projectile reaches its maximum height, you need to use the horizontal component of the projectile's motion. At the maximum height, the vertical velocity component is 0, and only the horizontal velocity component contributes to the distance from the starting point. The horizontal distance traveled is given by the formula:

$\text{Horizontal distance} = \text{horizontal velocity} \times \text{time}$

Given the initial velocity $v = \frac{8}{15}$ m/s and the launch angle $\theta = \frac{\pi}{12}$, you can find the horizontal velocity component ($v_x$) using:

$v_x = v \times \cos(\theta)$

Then, calculate the time it takes for the projectile to reach its maximum height using the vertical motion equation:

$t_{\text{max height}} = \frac{v_y}{g}$

Where $v_y$ is the initial vertical velocity component and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$).

Once you have $t_{\text{max height}}$, plug it into the horizontal distance formula to find the distance from the starting point when the projectile reaches its maximum height.